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Posted by on Wednesday, October 10, 2007 at 12:39pm.

A 250.0 kg roller coaster car has 20000 J of potential energy at the top of a hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?

  • Physical Science-please help - , Wednesday, October 10, 2007 at 12:46pm

    The KE will equal the PE at the top

    1/2 m v^2= 20000

    solve for v.

  • Physical Science-please help - , Wednesday, October 10, 2007 at 12:55pm

    Could you please put this in layman's terms?

  • Physical Science-please help - , Wednesday, October 10, 2007 at 1:05pm

    I believe that this answer is 9.8 m/s am I right?

  • Physical Science-please help - , Wednesday, October 10, 2007 at 1:29pm

    The kinetic energy (1/2 *mass*velocity squared) will equal the potential energy a the top

    1/2 m v2 = 20000
    The veloicity will be near 13 meters/second at the bottom

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